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I'm doing 1.2 in Lee's Introduction to smooth manifolds: Prove that the disjoint union of uncountably many copies of $\mathbb{R}$ is not second countable.

So first, let $I$ be the set over which we are unioning. Then I believe the disjoint union is just $\mathbb{R}\times I$. Then I believe that sets of the form $\cup_{x\in A}(x,i)$ is open if and only if $A\subset\mathbb{R}$ is open (I know the open sets are defined with the canonical injection though). At first I thought if I let $I=\mathbb{R}$, then $\mathbb{R}\times I=\mathbb{R^2}$, but now I am thinking maybe they just have the same elements, but the topology is different, and this is why $\mathbb{R}\times I$ is not second countable?

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More generally, you might like to prove that any topological space which has an uncountable family of pairwise disjoint open sets is not second countable. – Nate Eldredge Jul 13 '12 at 7:05
up vote 11 down vote accepted

It will be easier if we talk about $(0,1)$ which is homeomorphic to $\mathbb R$, but its diameter is $1$.

First note that the disjoint union is metrizable, by setting the distance between any two points coming from two different copies to be $2$.

For metric spaces second countability implies separability. However a countable subset of the disjoint union cannot meet all the copies, only countably many of them.

Therefore the disjoint union is not separable and thus not second countable.

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Just knowing that a space is a disjoint union of a collection of known subspaces is not enough to reconstruct the topology of the space. So when one talks about external disjoint unions, the standard convention is to take each of the unioned subspaces as an open set in the disjoint union.

In your case, this is equivalent to taking $I$ to be a discrete uncountable space, which does not produce a space homeomorphic to $\mathbb{R}^2$, but rather a space with uncountably many connected components, each of which is homeomorphic to $\mathbb{R}$.

At this point, you should be able to show why the space is not second countable on your own. As a hint, show that any basis has an element which is a subset of a copy of $\mathbb{R}$ for each copy of $\mathbb{R}$.

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Of the three arguments currently on offer, this is by far the most straightforward, I think. – Brian M. Scott Jul 13 '12 at 6:26
I do not understand how every basis can have for each disjoint copy of $\mathbb{R}$ an element that is its subset. Take the first disjoint copy $\mathbb{R}_1$, construct a basis $B$ for it. Then construct $B_2=\{U \times \mathbb{R}_2: U \in B\}$, similarly construct $B_3$. Then into the basis for the disjoint copies of $\mathbb{R}$ include $B_2$ and $B_3$, then clearly $B$ can be generated from $B_2$ and $B_3$ by taking the intersection, although none of the elements in the basis is the subset of $\mathbb{R}_1$. – Dávid Tóth Mar 2 '14 at 20:58
You are correct, I am wrong. I thought of a basis $\mathcal{B}$ for $\mathcal{T}$ as the minimal subset of $\mathcal{T}$ that can generate $\mathcal{T}$ with the operations of a union and a finite intersection. – Dávid Tóth Mar 6 '14 at 7:11

You're right, it is the fact that we are considering the disjoint union topology that makes it non-second-countable. While Asaf's proof is very slick, here's a (much longer) direct argument.

Let $A$ be an uncountable set, and let $X_\alpha=\mathbb{R}$ for each $\alpha\in A$. Let $X=\coprod_{\alpha\in A}X_\alpha$, with the disjoint union topology. The open sets of $X$ are those sets of the form $\coprod_{\alpha\in A}U_\alpha$ where $U_\alpha\subseteq X_\alpha$ is open. In particular, for any $\beta\in A$, $X_\beta=\coprod_{\alpha\in A}U_\alpha$ where $U_\beta=X_\beta$ and $U_\alpha=\emptyset$ for $\alpha\neq\beta$ is an open set of $X$.

Let $\mathcal{B}$ be any basis for the topology on $X$, so that every open set $U\subseteq X$ has $$U=\bigcup_{\substack{B\,\in\,\mathcal{B}\\B\,\subseteq\, U}} B.$$ Obviously, this implies that for any non-empty open set $U$ in $X$, there exists a non-empty $B\in\cal{B}$ such that $B\subseteq U$. In other words, for any non-empty open set $U$ in $X$, we have $|Z(U)|\geq 1$, where $$Z(U)=\{B\in\mathcal{B}\mid B\subseteq U,B\neq\emptyset\}.$$ Because each $X_\alpha$ is a non-empty open set in $X$, and $X_\alpha\cap X_\beta=\emptyset$ for any $\alpha\neq\beta$, we have that $$Z(X_\alpha)\cap Z(X_\beta)= \{B\in\mathcal{B}\mid B\subseteq X_\alpha\cap X_\beta,B\neq\emptyset\}=\emptyset\text{ for any }\alpha\neq\beta.$$

Because $\mathcal{B}\supseteq\bigcup_{\alpha\in A}Z(X_\alpha)$ and $Z(X_\alpha)\cap Z(X_\beta)=\emptyset$ for $\alpha\neq\beta$, we have $$|\mathcal{B}|\geq\sum_{\alpha\in A}|Z(U)|\geq\sum_{\alpha\in A}1=|A|,$$ Because $A$ is uncountable, $\mathcal{B}$ is uncountable. Thus, $X$ is not second-countable.

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Note that something stronger is true: No topological sum of uncountably many topological spaces is second countable.

Let $I$ be an uncountable set, and for each $i \in I$ let $X_i$ be some (nonempty) topological space. Without loss of generality we may assume that the $X_i$s are pairwise disjoint. Consider the topological sum $X = \bigoplus_{i \in I} X_i$. We show that $X$ is not second-countable.

Recall that $U \subseteq X$ is open in $X$ iff $U \cap X_i$ is open in $X_i$ for each $i \in I$. From this it follows that the family $$\mathcal{B} = \bigcup_{i \in I} \{ U \subseteq X_i : U \text{ is open in }X_i \}$$ is a base for $X$.

The following is a fairly easy fact to prove:

Given any base $\mathcal{D}$ for a topological space $Y$, there must be a base $\mathcal{D}^\prime$ for $Y$ of minimum possible size/cardinality included in $\mathcal{D}$.

Thus, if $X$ were second countable some countable subfamily of $\mathcal{B}$ must be a base for $X$. But the union of any countable subfamily of $\mathcal{B}$ can meet only countably many $X_i$s, and thus cannot be all of $X$!

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An attempt to simplify Arthur Fischer's argument:

Each copy of the base space is an open set, and these are pairwise disjoint. Since there uncountably many of them, the countable chain condition is not satisfied, therefore the union is not separable or second countable.

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